Re: crash tests, from a Physics Teacher

["David G" <daveglu@hotmail.com>]

> SO, for the 60/0 crash, we will need to take into account the amount of
> energy absorbed by the 30/30 car, and apply that to each of the 60/0 cars

 The momentum of the system becomes
1000kg(30m/s) + 1000kg(0m/s)  = 2000kg (15m/s)
which states that both vehicles (assuming inelastic collision) will be
traveling at 15m/s (30 mph) immediatly after the collision and the impact
becomes identical to the 30,-30 collision.
  What have I missed.  I have no intention of starting a torsen thread, I
just want to help solve poeples misconceptions especially if they are mine!
  Dave G